Chemistry - Concentration Terms Question with Solution | TestHub

ChemistryConcentration TermsConcentration termHard2 minPYQ_2012

ChemistryHardsingle choice

In the reaction of $KMnO_{4}$ with an oxalate in acidic medium, $MnO_{4}^{-}$ is reduced to $Mn^{2 +}$ and $C_{2} O_{4}^{2 -}$ is oxidised to $CO_{2}$ . Hence, $50 mL$ of 0.02 $M KMnO_{4}$ is equivalent to

Options:

Answer:

Solution:

eg. of $KMnO_{4}$ $= 5 \times 0.02 \times 50 = 250 \times 0.02 = 509.$ eq. of $KAnO_{4} \equiv eq_{1}$ of $C_{2} O_{4}_{2} -$ $5 = option 2 = 50 \times 0.05 \times 2 = 589.$

Stream:BITSATSubject:ChemistryTopic:Concentration TermsSubtopic:Concentration term

⏱ 2mℹ️ Source: PYQ_2012

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